pead, peadr

  
Purpose 		 

Compute the product of exponentials $\prod_{i=1}^{n}
e^{ad_{X_i}}X_{n+1}$.





Syntax 		
   e:=pead(n,max_bracket_order,X);


e:=peadr(n,max_bracket_order,X,B,lbt);




Description 		 

Computes the product of exponentials $\prod_{i=1}^{n}
e^{ad_{X_i}}X_{n+1}$.


peadr, reduces the Lie brackets in the expression
to elements in the PHB, and further simplifies the expression 
according to the supplied Lie bracket table.  If the Lie bracket table
is an empty list or set, no additional simplifications are carried
out.





Arguments 		 
 $n$  $\parbox{0.64\textwidth}{Number of products of
 exponentials.}$


 		 max_bracket_order


 		 		 $\parbox{0.64\textwidth}{The
 maximum bracket order in the exponentia...
...ion of
 $e^{ad_x}$; see \texttt{ead}, \texttt{eadr} on
 page~\pageref{ss:ead}.}$


 		 $X$ $\parbox{0.64\textwidth}{Label or name for the
 (indeterminate) eleme...
...ttt{peadr(n,max\_bracket\_order,B,B,lbt)}, where \texttt{B}
 contains the PHB.}$




 		 peadr additionally requires:


 		 $B$		 $\parbox{0.64\textwidth}{A Philip Hall basis.}$


 		 $lbt$		 $\parbox{0.64\textwidth}{A Lie bracket t...
...n empty list or set if no
 dependencies between the brackets in the PHB exist.}$




Examples 		

In this first example the computation of $\prod_{i=1}^2
e^{ad_{x_i}x_3}$ is considered. Here $x_i$ are Lie algebra
indeterminates, and the Taylor series expansion of the exponential
$ad$ operator is computed up to brackets of order two.
> pead(2,2,x);

  x[3] + g2 x[2] &* x[3] + 1/2 (g2 x[2] &* g2 x[2] &* x[3]) 
  
  + (g1 x[1] &* x[3]) + (g1 x[1] &* g2 x[2] &* x[3]) 

  + 1/2 (g1 x[1] &* (g2 x[2] &* g2 x[2] &* x[3]))

  + 1/2 (g1 x[1] &* (g1 x[1] &* x[3]))

  + 1/2 (g1 x[1] &* (g1 x[1] &* g2 x[2] &* x[3]))

  + 1/4 (g1 x[1] &* (g1 x[1] &* (g2 x[2] &* g2 x[2] &* x[3])))

For the next example consider the P. Hall basis B of the
example for the function phb on page .  
The computation of $\prod_{i=1}^2 e^{ad_{B_i}B_3}$ yields,
> pead(2,2,B);

  f2~ + (g2 f1~ &* f2~) + 1/2 (g2 f1~ &* (g2 f1~ &* f2~))

  + (g1 f0~ &* f2~) + (g1 f0~ &* (g2 f1~ &* f2~))

  + 1/2 (g1 f0~ &* (g2 f1~ &* (g2 f1~ &* f2~)))

  + 1/2 (g1 f0~ &* (g1 f0~ &* f2~))

  + 1/2 (g1 f0~ &* (g1 f0~ &* (g2 f1~ &* f2~)))

  + 1/4 (g1 f0~ &* (g1 f0~ &* (g2 f1~ &* (g2 f1~ &* f2~))))

Note that the above computation does not yields an expression whose
elements are in the P. Hall, if this additional simplification is
required then using peadr we obtain:
> peadr(2,2,B,B,{});

                                2
  f2~ + g2 (f1~ &* f2~) + 1/2 g2  (f1~ &* (f1~ &* f2~))

                                   2
         + g1 (f0~ &* f2~) + 1/2 g1  (f0~ &* (f0~ &* f2~))



		

Note that in the above example the degree of the highest order brackets is
three, unlike when pead was used, where the highest order
brackets are of degree five.   This is due to the fact that
peadr calls reduceLBT to make simplifications, thus
brackets in the series expansion whose order is higher than
max_bracket_order are eliminated.





See Also 		 

ad, ead, eadr.